Question: Solve for $x$ : $2x^2 + 20x + 50 = 0$
Dividing both sides by $2$ gives: $ x^2 + {10}x + {25} = 0 $ The coefficient on the $x$ term is $10$ and the constant term is $25$ , so we need to find two numbers that add up to $10$ and multiply to $25$ The number $5$ used twice satisfies both conditions: $ {5} + {5} = {10} $ $ {5} \times {5} = {25} $ So $(x + {5})^2 = 0$ $x + 5 = 0$ Thus, $x = -5$ is the solution.